Dynamics: Newton's Second Law

Tags:
1. Jan 8, 2015

yandereni

1. The problem statement, all variables and given/known data
Here's the problem:
A particular racecarcan cover a quartermile track (402m) in 6.40s starting from standstill. Assuming that the acceleration is constant, how many "g's"(all i know is that it means accleration due to gravity) does the driver experience? If the combined mass of the driver and the racecar is 485kg, what horizontal force must the road exert on the tires?

2. Relevant equations
1. Δy(displacement) = [(Velocityinitial)(Δt)] + ((1/2)acceleration)(Δt)2
2. Forcegrav + (acceleration)(mass)= Forcetotal
I'm not even sure if the second equation even exists but thats what i made out from the problem.
3. The attempt at a solution
I tried solving for the "g's" using the first equation and i substituted the possible things:

402m = [(0m/s)(6.4s)] + [((1/2)a)(6.4s)]
i don't know if this is the right equation but its the closest that i can get

then i tried solving for the horizontal force with the second equation and substituted the given:
[(-9.8m/s2)(485kg)] + [(19.6m/s2)(485kg)] = Forcetotal

and in case oyu are wondering where i got the acceleration(19.6m/s2) its from the first equation i solved (which i think is wrong)

-yandereni

2. Jan 8, 2015

Bandersnatch

Hi, yandereni.

You did well in the first part. It looks correct.

What are the directions of the two forces you've included here? What is the direction the question is concerned with?

3. Jan 8, 2015

yandereni

I
I looked at the problem again and there were no directions stated. And when i told my other classmates about the first part if i did it right, they said no and the answer should be 2.00m/s2.

4. Jan 8, 2015

Bandersnatch

what's this then, eh?
That's wrong. The car would cover some 41m at that acceleration.

5. Jan 8, 2015

Bandersnatch

Maybe they meant 2g, not 2m/s^2?

6. Jan 8, 2015

yandereni

oh, now i understand. but can i solve the horizontal force without any directions given?

7. Jan 8, 2015

Bandersnatch

But you do know which direction the car is moving, don't you? All you need to make sure when setting up your equation is not to include forces that are not acting (i.e., have no component) in the direction you're interested in.

8. Jan 8, 2015

yandereni

Thank you very much! I got the right answer and i get the concept now. Thanks a lot! :)